Question

are provided below. Parameters 1. Initial Cost ($) Case $330,000 John Deere 375,000 2. Revenues ($) $250,000 at EOY1 decreasing by 1% annually thereafter ІН 355,000 $225,000 at EOY1 increasing by 2% annually to EOY5 inclusively: revenues in ΕΟΥ5 subsequently increase annually by 1% to EOY10. $163,500 at EOY1 increasing by $500 annually thereafter $260,000 at EOY1 increasing by 1% annually thereafter. 3. Operating Costs $140,000 at EOY1 increasing by 3% annually thereafter. $190,000 from EOY1 to EOY5 inclusively; $200,000 from EOY6 to EOY10. $30,000 0 20,000 4. End-of-life salvage value ($) 5. Useful life (years) 5 years 10 years 10 years

NPW for IH

Answer 1

Net present value of IH is calculated below

					in $
Year	Outflow	inflow	Net	Discounting Factor @ 10%	Present Value
0	-3,55,000	-	-3,55,000	1.00	-3,55,000
1	-1,63,500	2,25,000	61,500	0.91	55,909
2	-1,64,000	2,29,500	65,500	0.83	54,132
3	-1,64,500	2,34,090	69,590	0.75	52,284
4	-1,65,000	2,38,772	73,772	0.68	50,387
5	-1,65,500	2,43,547	78,047	0.62	48,461
6	-1,66,000	2,45,983	79,983	0.56	45,148
7	-1,66,500	2,48,443	81,943	0.51	42,049
8	-1,67,000	2,50,927	83,927	0.47	39,153
9	-1,67,500	2,53,436	85,936	0.42	36,445
10	-1,68,000	2,55,971	87,971	0.39	33,916
					1,02,886